Is isset($foo) functionally identical to !$foo?

Question

Will isset($foo) always display the same result as !$foo?

I have a peice of code where I'm getting php warnings for using:

if(!$foo){}

And I'm pretty sure that I should be using:

if(!isset($foo)){}

And that made me curious whether I'm changing the functionality here or not.

Answer 1

No.

Using a boolean negation operator ! a variable is casted to boolean. Boolean FALSE is equal to NULL (this is functionally however the same as isset()), empty string, 0, empty array.

Using isset no error is given if the variable does not exist. If you use ! with non-existent variable, E_NOTICE is shown.

Answer 2

No.

One tests if a value is not set, the other tests if it is not true.

Compare:

<?php
$foo = 0;

if(!$foo){ echo 1; }
if(!isset($foo)){ echo 2; }
?>